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  • Show that for a first order reaction, time required for completion of 99% of reaction is twice the time required for complrtion of 90% of reaction.       <div style="

Show that for a first order reaction, time required for completion of 99% of reaction is twice the time required for complrtion of 90% of reaction.

 

 

 

 
 
 
 
 

Answers (1)

Time = \frac{2.303}{K}log \frac{initial concentration}{final concentration}

for 99% completion of reaction:

T_{1}= \frac{2.303}{K}\times \log \frac{100}{100-99}=\frac{2.303}{K}\times \log \frac{100}{1}= \frac{2 \times 2.303}{K}

for 90% completion of reaction:

T_{2}= \frac{2.303}{K}\times \log \frac{100}{100-90}=\frac{2.303}{K}\times \log 10= \frac{1 \times 2.303}{K}

\therefore T_{1} = 2\times T_{2}

Posted by

Sumit Saini

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